2022 USEMO #3

Point $P$ lies in the interior of a triangle $ABC$. Lines $AP$, $BP$, and $CP$ meet the opposite sides of triangle $ABC$ at $A$', $B'$, and $C'$ respectively. Let $P_A$ the midpoint of the segment joining the incenters of triangles $BPC'$ and $CPB'$, and define points $P_B$ and $P_C$ analogously. Show that if \[ AB'+BC'+CA'=AC'+BA'+CB' \]then points $P,P_A,P_B,$ and $P_C$ are concyclic. 

Let \(X_A\), \(X_B\), \(X_C\) be the projections of \(P_A\), \(P_B\), \(P_C\) onto \(\overline{BB'}\), \(\overline{CC'}\), \(\overline{AA'}\), respectively. Then\[PX_A=\tfrac14(BP+PC'-C'B)-\tfrac14(B'P+PC-CB'),\]so (with lengths appropriately directed) we have \(PX_A+PX_B+PX_C=0\). Let \(\alpha=\angle BPP_A\), and similarly define \(\beta\) and \(\gamma\), so \(\alpha+\beta+\gamma=90^\circ\). Let \(P_A'\), \(P_B'\), \(P_C'\) be the inverses of \(P_A\), \(P_B\), \(P_C\) about \(P\) with radius 1. Then\begin{align*}     \operatorname{Area}(\triangle P_A'P_B'P_C') &=\sum_\mathrm{cyc}\operatorname{Area}(\triangle PP_B'P_C') =\sum_\mathrm{cyc} PP_B'\cdot PP_C'\cdot\sin(\alpha+\beta)\\ &=\sum_\mathrm{cyc}\frac{\cos\alpha}{PX_B}\cdot\frac{\cos\beta}{PX_C}\cdot\cos\gamma =\frac{\cos\alpha\cos\beta\cos\gamma}{PX_A\cdot PX_B\cdot PX_C}\cdot\sum_\mathrm{cyc} PX_A=0. \end{align*}It follows that \(P_A'\), \(P_B'\), \(P_C'\) are collinear, as desired. $\square$


Popular posts from this blog

1995 IMO #2

Inequality problem I made